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2 years ago

What is the solution to the equation 13.7y = 6.2y + 30

Mathematics

1 answer:

andre [41]2 years ago

6 0

Answer:

Step-by-step explanation:

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HURRY IM DOING AN ALGEBRA TEST!?!!!

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Answer:

True?

Step-by-step explanation:

I think its true but I'm not 100% sure so I would go with true. Try thinking of your past problems that you used and think about that maybe? But in then mean time, I would go with True

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2 years ago

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How can individual attitudes and lack of response to injustice harm society as a whole? (this was meant to be in history)

dmitriy555 [2]

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2 years ago

the length of a rectangular parking lot at the airport is 2/3 is mile.if the area is 1/2 square mile,what is the width of the pa

ch4aika [34]

Answer:

3/4

Step-by-step explanation:

3/4*2/3+1/2

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2 years ago

Solve log 4x + log 13 = 0. Round to the nearest thousandth if necessary. a.0.019 b. 52 c.3.25 d. 0.311 Mathematics

kumpel [21]

Log4 + logX + log13 = 0
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2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago